How strong must the magnetic field

Question

A be‎am of ele‎ctrons (of charge q_e and mass m_e) is s‎hot into a reg‎ion of unif‎orm magn‎etic fi‎eld with a spee‎d of 513[m/s]. Th‎ey exit the re‎gion o‎f uni‎form magn‎etic fi‎eld op‎posite the dire‎ction they had ente‎red. How stron‎g must the m‎agnetic field be s‎uch that ea‎ch elec‎tron stay‎ed in the regi‎on for a total of 815[ms]

Answer 1

In the absence of any information to the contrary, I will assume that the electrons

are shot in a direction perpendicular to the magnetic field.

Inside the magnetic field each electron will then follow a circular trajectory

in a plane perpendicular to the magnetic field.

The centripetal force responsible for the circular motion is created by the

action of the magnetic field on the electrons.

Inside the magnetic field each electron follows a half circle before exiting.

Let

r (unknown) be the radius of the trajectory,

F (unknown) be the magnitude of the centripetal force,

B (to be calculated) be the strength of the magnetic field,

q = 1.6×10^-19 C be the charge of an electron,

m = 9.1×10^-31 kg be the mass of an electron,

v = 513 m/s be the speed,

t = 815 ms = 0.815 s be the time the electron spends in the magnetic field,

s (unknown) be the length of the half-circle traversed by the electron.

The above quantiles are related by the following equations:

F = q(v×B) (by Lorentz law)

F = mv²/r (by Newton's second law applied to circular motion)

s = πr (length of the half circle)

s = vt (by definition of speed)

From the last two equations

r = vt/π

From the first two equations

q(v×B) = mv²/r = πmv/t

By our assumption that v and B are perpendicular we can rewrite it as

qvB = πmv/t

B = πm/qt

(Note that the result is independent of the speed v.)

Substituting actual numbers

B = π×9.1×10^-31/(1.6×10^-19×0.815) = 2.2×10^-11 T